Minimum and maximum Distance of a point from an ellipse We have to find the minimum and maximum distance of a point $(9/5 ,12/5)$ from the ellipse $4(3x+4y)^2 +9(4x-3y)^2 =900$ .
I got to know the center of ellipse would be $(0,0)$ .
after that I am not getting any start.
I think there would be some trick to solve it .
 A: Introduce the following variables (Leibovici style$^*$)
$$u=2(3x+4y)\,\,\text{ and } \,\, v=3(4x-3y).$$
Then we get a circle of radius $30$ in the $u,v$ coordinate system:$$u^2+v^2=900.$$
In this coordinate system the point from which we measure the distance of the points on the circle is
$$(30, 0).$$
So, the closest point on the circle to this point is at $(30,0)$ and the farthest point on the circle is at $(-30,0)$. 
The minimum distance is $0$ and the corresponding point on the ellipse is $\left(\frac95,\frac{12}5\right)$.
For the other (farthest) point, use the inverse of the transformation we used to get the coordinates of the point $(-30,0)$ points in the $(x,y)$ system. Or solve the following system of equations
$$-30=6x+8y\ \ \\  \ \ \ \ \ 0=12x-9y.$$
Then calculate the distance of the two points that we have just found.

EDIT
The solution is $\left(-\frac95,-\frac{12}5\right)$. The closest point is $\left(\frac95,\frac{12}5\right)$. The distance is
$$\sqrt{\left(\frac{18}5\right)^2+\left(\frac{24}5\right)^2}=6.$$

$^*$Claude Leibovici suggested in a later deleted comment the transformation
$$u=3x+4y\,\,\text{ and } \,\, v=4x-3y.$$
A: 1) That formula is kinda hard to work with. You should start by rewriting it to collect all the $x$ terms together and then collect all the $y$ terms together.
2) the simplest way to solve problems like this tends to be basic calculus! Write down the formula for the distance from a point on the ellipse to your point, and differentiate!
A: Hint...try rotating the ellipse by using $u=\frac{3x+4y}{5}$ and $v=\frac{4x-3y}{5}$ and transforming the given fixed point
